Question

The fourth vertex D of a parallelogram ABCD whose three vertices are A\(-4, 1\), B\(4, 5\) and C\(6, 1\) is :

• A. \(-2, -3\)
• B. \(3, -2\)
• C. \(0, -1\)
• D. \(0, 1\)

Hint:

For any parallelogram \(ABCD\) in order, the sum of coordinates of opposite vertices is equal because of the midpoint property:
\[ x_A + x_C = x_B + x_D \implies x_D = x_A + x_C - x_B \]
\[ y_D = y_A + y_C - y_B \]
Let's check this shortcut:
\[ x_D = -4 + 6 - 4 = -2 \]
\[ y_D = 1 + 1 - 5 = -3 \]
This gives the coordinates \((-2, -3)\) in seconds without needing midpoint formulas.

Answer 1

The Correct Option is A

Step 1: Understand what is being asked.
We are given three vertices $A(-4,1)$, $B(4,5)$, $C(6,1)$ of a parallelogram $ABCD$ and need the fourth vertex $D$. Instead of using the midpoint of diagonals, let us use the fact that opposite sides of a parallelogram are equal and parallel.

Step 2: Use the parallel-side property.
In parallelogram $ABCD$, side $AB$ is parallel to and equal in length to side $DC$, and both point in the same direction when we go from $A$ to $B$ and from $D$ to $C$. This means:
\[ \overrightarrow{AB} = \overrightarrow{DC} \]
Step 3: Write this condition in coordinates.
$\overrightarrow{AB}$ is the change in coordinates going from $A$ to $B$:
\[ \overrightarrow{AB} = (4-(-4),\ 5-1) = (8,4) \]
Let $D=(x,y)$. Then $\overrightarrow{DC}$ is the change in coordinates going from $D$ to $C$:
\[ \overrightarrow{DC} = (6-x,\ 1-y) \]
Step 4: Equate and solve for $D$.
Since $\overrightarrow{AB}=\overrightarrow{DC}$:
\[ 6-x = 8 \implies x = 6-8 = -2 \]
\[ 1-y = 4 \implies y = 1-4 = -3 \]
Final Answer:
The fourth vertex $D$ is $(-2,-3)$, matching option (A).
\[ \boxed{D(-2,-3)} \]